intersections of the angle bisector with intouch triangle

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intersections of the angle bisector with intouch triangle

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Source: Middle European Mathematical Olympiad T-5

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#1 h Sep 21, 2014, 2:11 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with $AB < AC$ . Its incircle with centre $I$ touches the sides $BC, CA,$ and $AB$ in the points $D, E,$ and $F$ respectively. The angle bisector $AI$ intersects the lines $DE$ and $DF$ in the points $X$ and $Y$ respectively. Let $Z$ be the foot of the altitude through $A$ with respect to $BC$ .

Prove that $D$ is the incentre of the triangle $XYZ$ .

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jayme

#2 h Sep 21, 2014, 3:33 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
an outline of my proof...

1. M the midpoint of BC
2. the triangle MXY is M-isoceles; ZM is the Z-inner bissector of the triangle ZXY
3. M is the center of the circumcircle of DXY (Morel's circle) (see http://jl.ayme.pagesperso-orange.fr/ vol. 4 Symetrique de (OI) p. 5)
4. According to the A-Mention circle (Shamrock theorem), D is the incenter of ZXY.

Sincerely
Jean-Louis

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#3 h Sep 24, 2014, 9:22 PM • 2 Y

Y by Adventure10, Mango247

See attachment.

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#4 h Sep 25, 2014, 6:44 AM • 1 Y

Y by Adventure10

We have $\angle YAF=A/2$ and $\angle YFA=90^0+(C/2)$ .Hence $\angle DYX=\angle AYF=(B/2)$ .Also $\angle IBD=(B/2)$
Hence $IBDY$ is cyclic.Now $ID\perp BD \implies IY\perp BY$ since $IBDY$ is cyclic.Also $AZ\perp BZ$ ,hence $ABZY$ is cyclic.
Hence $\angle YZC=\angle BAY=(A/2)$ .Similarly we can prove that $\angle XZC=(A/2)$ .Hence $DZ$ bisects $\angle XZY$ .
Now $\angle BDY=90^0+(C/2)$ .Hence $\angle ZYD=180^0-(90^0+(C/2))-(A/2)=(B/2)=\angle DYX$ .Hence $DY$ bisects $\angle ZYX$ .Hence $D$ is the incentre of $\triangle XYZ$ .

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Mikasa

#5 h Sep 25, 2014, 8:56 AM • 3 Y

Y by PatrikP, Adventure10, Mango247

Let $P\in AC$ such that $AP=AB$ . Let $BP\cap AI=X_0$ . Let $EX_0$ meet $\odot DEF$ at $D_0\neq E$ . Clearly $D_0$ is on the other side of $BP$ than $F$ . Now, $AI$ is the perpendicular bisector of $EF$ and $BP$ . Again, $\angle FD_0X_0=\angle ED_0F=\angle AFE=\angle ABP=\angle FBX_0$ i.e. $D_0BFX_0$ is cyclic. Thus, $\angle BDF=\angle BX_0F=90^{\circ}-\angle FX_0A=\angle X_0FE=\angle X_0EF=\angle DEF$ . So, $BD_0$ is tangent to $\odot DEF$ . This means that $D\equiv D_0$ . Thus $X\equiv X_0$ . So we proved that $BX\perp AI$ .

Now, $\angle DYE=\angle FYE=180^{\circ}-\angle YFE-\angle YEF$ . But $Y$ is on the perpendicular bisector of $EF$ . So, $\angle DYE=180^{\circ}-2\angle YFE=180^{\circ}-2\angle DFE=180^{\circ}-2\angle DEC=\angle DCE$ . Thus $DECY$ is cyclic, and so $\angle IYE=\dfrac{1}{2}\angle DYE=\dfrac{1}{2}\angle DCE=\angle ICE$ which means $IYCE$ is cyclic too. Thus $\angle IYC=\angle IEC=90^{\circ}$ . So $CY\perp AI$ too.

Now, $AXZB$ and $AZYC$ both are cyclic. So, $\angle ZXY=90^{\circ}-\angle BXZ=90^{\circ}-\angle BAZ=\angle ABC$ and $\angle DXY=\angle AXE=90^{\circ}-\angle DEF=90^{\circ}-\angle BDF=\angle IBD=\dfrac{1}{2}\angle ABC$ . This means $DX$ bisects $\angle ZXY$ . Also, $\angle XZD=180^{\circ}-\angle XZB=\angle XAB=\angle IAB=\angle IAC=\angle YAC=\angle YZC$ $=\angle YZD$ . Thus $DZ$ bisects $\angle XZY$ .

All these results imply that $D$ is the incenter of $\triangle XYZ$ .

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#6 h Jan 27, 2015, 8:31 AM • 3 Y

Y by godofgeometry, mahditorogh, Adventure10

Hi
Here's my solution.
1)By angle chasing we understand that DXY=B/2 and DYX=C/2 and ZDY=90+B/2
2)If we prove that ZXD=B/2 the problem will be solved.
3)So we need to say that ZXY=B.
4)Because IBD=B/2 so IBDX is cyclic. So XBD=XID but because DI and AZ are parallel we have XID=XAZ so ABZX is cyclic.
5)So ZXY=B so we are done.

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#7 h Jan 29, 2015, 9:50 PM • 1 Y

Y by Adventure10

EDIIT...

This post has been edited 1 time. Last edited by dothef1, Jan 17, 2016, 10:15 AM
Reason: false

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#8 h Sep 24, 2020, 6:48 PM • 2 Y

Y by Mango247, Mango247

$\angle IXD=180-\tfrac {\angle A}{2} -\angle AED=90-\tfrac {\angle A}{2}-\tfrac {\angle B}{2}=\tfrac {\angle C}{2}$
So $IDXE$ is cyclic.
$\angle IXC=180-\angle IFC=90=\angle AZC$
So $AZXC$ is cyclic as well. And now,
$\angle DXZ=\angle EDB-\angle CZX=180-\angle AEX-\tfrac{\angle A}{2}=\angle DXY$
Similarly we show $YD$ bisects $\angle XYZ$ and so $I$ is the incenter $XYZ$

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